a. Three processes share four resource units that can be reserved and released only one at a time. Each process needs a maximum of two units. Show that a deadlock cannot occur. b. N processes share M resource units that can be reserved and released only one at a time. The maximum need of each process does not exceed M, and the sum of all maximum needs is less than M + N. Show that a deadlock cannot occur.
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- 22. A system has two processes and three identical resources. Each process needs a maximum of two resources. Is deadlock possible? Explain your answer 23. Consider the previous problem again, but now with p processes each needing a maximum of m resources and a total of r resources available. What condition must hold to make the system deadlock free? please answer the second question...assume a system has 6 identical resources and N processes competing for them. each process can request at most 2 resources. what is the maximum value of N for the system to be deadlock free?39. A system has 'm' number of resources of same type and 3 processes A, B, C. Share these resources A, B, C which have the peak demand of 3, 4 and 6 respectively. Deadlock will not occur if the value of 'm' is __________. a. m = 15 b. m = 8 c. m = 13 d. m = 9
- 38. Consider a computer system with 6 tape drives and 'n' processes completing for them. What is the maximum value of 'n' for the system to be deadlock free? (Assuming that each processes may need 3 tape drives) a. 3 b. 2 c. 4 d. 7We are dealing with 6 processes p1;p2; : : : ;p6:(a) Processes p1;p2 in p3 arrive at time 2 ms, respectively. Each of those processesneeds 5 ms of CPU time.(b) Process p4 arrives at time 10 ms and needs 3 ms of CPU time.(c) Processes p5 in p6 arrive at time 19 ms. Process p5 needs 7 ms, while process p6needs 4 ms of CPU time.Scheduling is preemptive, where there process switch takes 1 ms. For the Round-Robinalgorithms with quantum 2 ms illustrate the execution of those processes on a timeline.Then determine the average turnaround time.Consider three processes (process id 0, 1, 2 respectively) with computing bursts timeunits of 3, 4 and 7 respectively. All processes arrive at the same time. Consider thescheduling algorithm with the longest remaining time first (LRTF). In LRTF ties arebroken by giving priority to the process with the lowest process id. What is theaverage waiting time for the three processes? Describe in depth.
- At times t = 0, 1, and 2 seconds, processes P1, P2, and P3 arrive in the Ready state, respectively. Suppose their execution times are 5 seconds, 4 seconds, and 3 seconds, respectively. Show how they will be scheduled using round robin scheduling with a quantum of 3 seconds. Calculate the average turnaround time per process.Consider a system consisting of m resources of the same type being shared by n processes, n > m. Each process has a maximum need of m/2 resources. Initially, each process has no resource requests. A process can request or release only one resource at a time. With n=3k and m=2k, for some integer k, show that the system is deadlock free.Consider a swapping system in which memory consists of the following hole sizes inmemory order: 12 MB, 4 MB, 20 MB, 12 MB,18 MB, 7 MB, 9 MB and 12 MB.Which hole is taken for successive segment requests of(a) 10MB(b) 15MB(c) 9 MBfor first fit, worst fit, and next fit?
- The following processes are being scheduled using a preemptive, priority-based, round-robin scheduling algorithm. Process priority Burst Arrival P1 8 15 0 P2 3 20 0 P3 4 20 20 P4 4 20 25 P5 5 5 45 P6 5 15 55 Each process is assigned a numerical priority, with a…In your opinion what does the term semaphore mean in terms of managing concurrent processes in operating systems. a) consider the case of a non negative counting semaphore S. During an execution, 18P(S) operations and 13 V (S) operations were carried out in some order. What is the largest initial value of S for which at least two P(S) operations are blocked?answer for part 1 is provided.I need answer for part 3 context switching At arrival time 0, we have 2 processes P1 and P2 with same priority 3 So, according to Shortest Remaining Time First policy, P1 with burst time of 15 P1 0 15 After 15 secs, no other process has arrived and Process P2 is executed for 5 secs and then new processes has arrived. But, Process P2 has the highest priority when compared to the new processes. So, it completely gets executed P1 P2 0 15 35 Now, all the processes have arrived and P3 P4 P5 has the same priority. So, according to SRTF policy, P5 is executed and then P4 and P3 P1 P2 P5 P3 P4 0 15 35 40 50 70 Now, the only remained process of P6 is executed P1 P2 P5 P3 P4 P6 0 15 35 40 50 70 85 Step 2 Thus, the…